餐饮哲学家问题 - 只有 2 个线程工作

Dining philosophers problem - only 2 thread worked

本文关键字：线程 工作 只有 哲学家 问题      更新时间：2023-10-16

我正在尝试解决餐饮哲学家的问题。

就我而言，每个哲学家都应该吃100万次。 问题是它似乎只有"1"并且"3"吃完了。 我正在使用具有关键部分锁定的线程，这是我的代码：

CRITICAL_SECTION ghCARITICALSection1;
CRITICAL_SECTION ghCARITICALSection2;
CRITICAL_SECTION ghCARITICALSection3;
CRITICAL_SECTION ghCARITICALSection4;
CRITICAL_SECTION ghCARITICALSection5;
DWORD WINAPI func(int* phiphilosopher)
{
if (1 == *phiphilosopher && TryEnterCriticalSection(&ghCARITICALSection1) && TryEnterCriticalSection(&ghCARITICALSection2))
{
std::cout << "1 is eating...n";
for (int i = 0; i < 1000000; i++)
{
i = i;
}
LeaveCriticalSection(&ghCARITICALSection1);
LeaveCriticalSection(&ghCARITICALSection2);
}
if (2 == *phiphilosopher && TryEnterCriticalSection(&ghCARITICALSection2) && TryEnterCriticalSection(&ghCARITICALSection3))
{
std::cout << "2 is eating...n";
for (int i = 0; i < 1000000; i++)
{
}
LeaveCriticalSection(&ghCARITICALSection2);
LeaveCriticalSection(&ghCARITICALSection3);
}
if (3 == *phiphilosopher && TryEnterCriticalSection(&ghCARITICALSection3) && TryEnterCriticalSection(&ghCARITICALSection4))
{
std::cout << "3 is eating...n";
for (int i = 0; i < 1000000; i++)
{
}
LeaveCriticalSection(&ghCARITICALSection3);
LeaveCriticalSection(&ghCARITICALSection4);
}
//...also for 4,5
return 0;
}

int philosopher1 = 1;
int* philosopher1ptr = &philosopher1;
int philosopher2 = 2;
int* philosopher2ptr = &philosopher2;
//...Also for philosopher 3,4,5
InitializeCriticalSection(&ghCARITICALSection1);
InitializeCriticalSection(&ghCARITICALSection2);
//...aslo for ghCARITICALSection 3,4,5
HANDLE WINAPI th1 = CreateThread(NULL, 0, (LPTHREAD_START_ROUTINE)func, philosopher1ptr, 0, NULL);
HANDLE WINAPI th2 = CreateThread(NULL, 0, (LPTHREAD_START_ROUTINE)func, philosopher2ptr, 0, NULL);
////...aslo for th3,4,5
WaitForSingleObject(th1, INFINITE);
WaitForSingleObject(th2, INFINITE);
//...also for th3,4,5
  • 每个哲学家都必须交替思考和吃饭。然而，哲学家只有在左右叉子都有时才能吃意大利面。每个叉子只能由一个哲学家握住，因此一个哲学家只有在另一个哲学家不使用叉子的情况下才能使用叉子。

想想这里的逻辑

if (TryEnterCriticalSection(&a) && TryEnterCriticalSection(&b)) {
// . . .
LeaveCriticalSection(&a);
LeaveCriticalSection(&b);
}

如果TryEnterCriticalSection(&a)成功而TryEnterCriticalSection(&b)失败会发生什么;CSa永远处于进入状态。

它应该看起来像

if (TryEnterCriticalSection(&a)) {
if (TryEnterCriticalSection(&b)) {
// . . .
LeaveCriticalSection(&b);
}
LeaveCriticalSection(&a);
}
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